Assignment 4

Data wrangling

Data analysis

Load the Boston data from the MASS package. Explore the structure and the dimensions of the data and describe the dataset briefly, assuming the reader has no previous knowledge of it. Details about the Boston dataset can be seen for example here. (0-1 points)

# access the MASS package
library(MASS)

# load the data
data(Boston)

#Explore the structure and the dimensions of the data
dim(Boston)

## [1] 506  14

str(Boston)

## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...

The Boston data frame has 506 rows and 14 columns.

This dataset contains information collected by the U.S Census Service concerning housing in the area of Boston Mass.

Details of the dataset: crim : per capita crime rate by town.

zn : proportion of residential land zoned for lots over 25,000 sq.ft.

indus : proportion of non-retail business acres per town.

chas :Charles River dummy variable (= 1 if tract bounds river; 0 otherwise).

nox : nitrogen oxides concentration (parts per 10 million).

rm : average number of rooms per dwelling.

age :proportion of owner-occupied units built prior to 1940.

dis :weighted mean of distances to five Boston employment centres.

rad :index of accessibility to radial highways.

tax :full-value property-tax rate per $10,000.

ptratio :pupil-teacher ratio by town.

black : 1000(Bk - 0.63)^21000(Bk−0.63) 2 where Bk is the proportion of blacks by town.

lstat : lower status of the population (percent).

medv : median value of owner-occupied homes in $1000s.

3. Show a graphical overview of the data and show summaries of the variables in the data. Describe and interpret the outputs, commenting on the distributions of the variables and the relationships between them. (0-2 points)

summary(Boston)

##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08205   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

# access the GGally and ggplot2 libraries
library(GGally)

## Loading required package: ggplot2

## Registered S3 method overwritten by 'GGally':
##   method from   
##   +.gg   ggplot2

library(ggplot2)
pairs(Boston)

I think this looks ugly and I can not see anything. Lets try something else.

p1 <- ggplot(Boston) + 
  geom_histogram(aes(x = crim), fill = "#E69F00") +
labs(title="per capita crime rate by town")

p2 <- ggplot(Boston) + 
    geom_histogram(aes(x = zn), fill = "#56B4E9") +
  labs(title="proportion of residential land zoned for lots over 25,000 sq.ft.")

p3 <- ggplot(Boston) + 
    geom_histogram(aes(x = indus), fill = "#009E73")+
  labs(title="proportion of non-retail business acres per town")

p4 <- ggplot(Boston) + 
    geom_histogram(aes(x = chas), fill = "#F0E442")+
  labs(title="Charles River dummy variable")

p5 <- ggplot(Boston) + 
    geom_histogram(aes(x = nox), fill = "#0072B2")+
  labs(title="nitrogen oxides concentration")

p6 <- ggplot(Boston) + 
  geom_histogram(aes(x = rm), fill = "#D55E00")+
  labs(title="average number of rooms per dwelling.")

p7 <- ggplot(Boston) + 
  geom_histogram(aes(x = age), fill = "wheat1")+
  labs(title="proportion of owner-occupied units built prior to 1940")


p8 <- ggplot(Boston) + 
  geom_histogram(aes(x = dis), fill = "green")+
  labs(title="weighted mean of distances to five Boston employment centres")

p9 <- ggplot(Boston) + 
  geom_histogram(aes(x = rad ), fill = "mediumorchid2")+
  labs(title="index of accessibility to radial highways")

p10 <- ggplot(Boston) + 
  geom_histogram(aes(x = tax  ), fill = "grey")+
  labs(title="full-value property-tax rate per $10,00")

p11 <- ggplot(Boston) + 
  geom_histogram(aes(x = ptratio  ), fill = "pink4")+
  labs(title="pupil-teacher ratio by town")

p12 <- ggplot(Boston) + 
  geom_histogram(aes(x = black  ), fill = "black")+
  labs(title="1000(Bk−0.63)2 where Bk is the proportion of blacks by town.")

p13 <- ggplot(Boston) + 
  geom_histogram(aes(x = lstat  ), fill = "seagreen2")+
  labs(title="lower status of the population")


p14 <- ggplot(Boston) + 
  geom_histogram(aes(x = medv  ), fill = "coral")+
  labs(title="median value of owner-occupied homes in $1000s.")

library(gridExtra)

gridExtra::grid.arrange(p1, p2, p3, p4, nrow=2)

## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

library(gridExtra)

gridExtra::grid.arrange(p5, p6, p7, p8, nrow=2)

## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

library(gridExtra)

gridExtra::grid.arrange(p9, p10, p11, nrow=2)

## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

library(gridExtra)

gridExtra::grid.arrange(p12, p13, p14, nrow=2)

## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.

# calculate the correlation matrix and round it
cor_matrix <- cor(Boston) %>% round(digits = 2)

# print the correlation matrix
cor_matrix

##          crim    zn indus  chas   nox    rm   age   dis   rad   tax ptratio
## crim     1.00 -0.20  0.41 -0.06  0.42 -0.22  0.35 -0.38  0.63  0.58    0.29
## zn      -0.20  1.00 -0.53 -0.04 -0.52  0.31 -0.57  0.66 -0.31 -0.31   -0.39
## indus    0.41 -0.53  1.00  0.06  0.76 -0.39  0.64 -0.71  0.60  0.72    0.38
## chas    -0.06 -0.04  0.06  1.00  0.09  0.09  0.09 -0.10 -0.01 -0.04   -0.12
## nox      0.42 -0.52  0.76  0.09  1.00 -0.30  0.73 -0.77  0.61  0.67    0.19
## rm      -0.22  0.31 -0.39  0.09 -0.30  1.00 -0.24  0.21 -0.21 -0.29   -0.36
## age      0.35 -0.57  0.64  0.09  0.73 -0.24  1.00 -0.75  0.46  0.51    0.26
## dis     -0.38  0.66 -0.71 -0.10 -0.77  0.21 -0.75  1.00 -0.49 -0.53   -0.23
## rad      0.63 -0.31  0.60 -0.01  0.61 -0.21  0.46 -0.49  1.00  0.91    0.46
## tax      0.58 -0.31  0.72 -0.04  0.67 -0.29  0.51 -0.53  0.91  1.00    0.46
## ptratio  0.29 -0.39  0.38 -0.12  0.19 -0.36  0.26 -0.23  0.46  0.46    1.00
## black   -0.39  0.18 -0.36  0.05 -0.38  0.13 -0.27  0.29 -0.44 -0.44   -0.18
## lstat    0.46 -0.41  0.60 -0.05  0.59 -0.61  0.60 -0.50  0.49  0.54    0.37
## medv    -0.39  0.36 -0.48  0.18 -0.43  0.70 -0.38  0.25 -0.38 -0.47   -0.51
##         black lstat  medv
## crim    -0.39  0.46 -0.39
## zn       0.18 -0.41  0.36
## indus   -0.36  0.60 -0.48
## chas     0.05 -0.05  0.18
## nox     -0.38  0.59 -0.43
## rm       0.13 -0.61  0.70
## age     -0.27  0.60 -0.38
## dis      0.29 -0.50  0.25
## rad     -0.44  0.49 -0.38
## tax     -0.44  0.54 -0.47
## ptratio -0.18  0.37 -0.51
## black    1.00 -0.37  0.33
## lstat   -0.37  1.00 -0.74
## medv     0.33 -0.74  1.00

# visualize the correlation matrix
library(corrplot)

## corrplot 0.92 loaded

corrplot(cor_matrix, method="circle", type = "upper", cl.pos = "b", tl.pos = "d",  tl.cex = 0.6)

4. Standardize the dataset and print out summaries of the scaled data. How did the variables change? Create a categorical variable of the crime rate in the Boston dataset (from the scaled crime rate). Use the quantiles as the break points in the categorical variable. Drop the old crime rate variable from the dataset. Divide the dataset to train and test sets, so that 80% of the data belongs to the train set. (0-2 points)

# center and standardize variables
boston_scaled <- scale(Boston) 

# summaries of the scaled variables

summary(boston_scaled)

##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv        
##  Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 3.5453   Max.   : 2.9865

# class of the boston_scaled object
class(boston_scaled)

## [1] "matrix" "array"

# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)

boston_scaled$crim <- as.numeric(boston_scaled$crim)
summary(boston_scaled$crim)

##      Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
## -0.419367 -0.410563 -0.390280  0.000000  0.007389  9.924110

# create a quantile vector of crim and print it
bins <- c(quantile(boston_scaled$crim))
bins

##           0%          25%          50%          75%         100% 
## -0.419366929 -0.410563278 -0.390280295  0.007389247  9.924109610

# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels =c("low", "med_low", "med_high", "high"))

# look at the table of the new factor crime
table(crime)

## crime
##      low  med_low med_high     high 
##      127      126      126      127

library(tidyverse)

## ── Attaching packages ─────────────────────────────────────── tidyverse 1.3.2 ──
## ✔ tibble  3.1.8      ✔ dplyr   1.0.10
## ✔ tidyr   1.2.1      ✔ stringr 1.4.1 
## ✔ readr   2.1.2      ✔ forcats 0.5.2 
## ✔ purrr   0.3.4      
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::combine() masks gridExtra::combine()
## ✖ dplyr::filter()  masks stats::filter()
## ✖ dplyr::lag()     masks stats::lag()
## ✖ dplyr::select()  masks MASS::select()

boston_scaled <- boston_scaled %>% 
mutate(crime = crime)

# remove the crime variable from test data
boston_scaled <- dplyr::select(boston_scaled, -crim)

# number of rows in the Boston dataset 
n <- nrow(boston_scaled)

# choose randomly 80% of the rows
ind <- sample(n,  size = n * 0.8)

# create train set
train <- boston_scaled[ind,]

# create test set 
test <- boston_scaled[-ind,]

5. Fit the linear discriminant analysis on the train set. Use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables. Draw the LDA (bi)plot. (0-3 points)

# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)

# print the lda.fit object
lda.fit

## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2425743 0.2500000 0.2574257 0.2500000 
## 
## Group means:
##                   zn      indus        chas        nox         rm        age
## low       1.00580296 -0.9219706 -0.07145661 -0.8780561  0.4312278 -0.9066377
## med_low  -0.08203002 -0.2929945 -0.03844192 -0.5978736 -0.1081540 -0.3682561
## med_high -0.37839823  0.1621987  0.14409500  0.3647508  0.1172816  0.3713807
## high     -0.48724019  1.0171306 -0.11640431  1.0746907 -0.4729992  0.7986659
##                 dis        rad        tax    ptratio      black        lstat
## low       0.8669891 -0.6783478 -0.7632450 -0.4493788  0.3789822 -0.762077574
## med_low   0.3949980 -0.5565973 -0.4616871 -0.1152884  0.3101465 -0.125892185
## med_high -0.3483200 -0.4131592 -0.3245498 -0.2663750  0.1086021 -0.009962127
## high     -0.8541043  1.6379981  1.5139626  0.7806252 -0.7899197  0.966049643
##                 medv
## low       0.53087360
## med_low   0.00762892
## med_high  0.19329678
## high     -0.76973588
## 
## Coefficients of linear discriminants:
##                   LD1         LD2         LD3
## zn       0.1067523648  0.82508772 -1.04676538
## indus    0.0002587703 -0.25284291  0.20010146
## chas    -0.1020905719 -0.04108882  0.06388783
## nox      0.4270044165 -0.81916732 -1.33105280
## rm      -0.0964848210 -0.16805003 -0.05538190
## age      0.1934183051 -0.26457038 -0.12914859
## dis     -0.0675534871 -0.40355592  0.36860853
## rad      3.2762344817  1.11515180 -0.34181720
## tax     -0.1210357072 -0.22135539  1.04162236
## ptratio  0.1223561625 -0.05173073 -0.37841771
## black   -0.1471169791  0.06097538  0.13101189
## lstat    0.2620839171 -0.24317556  0.42421258
## medv     0.1693946904 -0.49870420 -0.14346248
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9447 0.0400 0.0153

# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# target classes as numeric
classes = as.numeric(train$crime)


# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)

6. Save the crime categories from the test set and then remove the categorical crime variable from the test dataset. Then predict the classes with the LDA model on the test data. Cross tabulate the results with the crime categories from the test set. Comment on the results. (0-3 points)

# save the correct classes from test data
correct_classes <- test$crime

# remove the crime variable from test data
test <- dplyr::select(test, -crime)

# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)

# cross tabulate the results
table(correct = correct_classes , predicted = lda.pred$class)

##           predicted
## correct    low med_low med_high high
##   low       14      12        3    0
##   med_low    4      16        5    0
##   med_high   0       6       15    1
##   high       0       0        0   26

7. Reload the Boston dataset and standardize the dataset (we did not do this in the Exercise Set, but you should scale the variables to get comparable distances). Calculate the distances between the observations. Run k-means algorithm on the dataset. Investigate what is the optimal number of clusters and run the algorithm again. Visualize the clusters (for example with the pairs() or ggpairs() functions, where the clusters are separated with colors) and interpret the results. (0-4 points)

# access the MASS package
library(MASS)

# load the data
data(Boston)

# center and standardize variables
boston_scaled <- scale(Boston) 

# euclidean distance matrix
dist_eu <- dist(Boston)

# look at the summary of the distances
summary(dist_eu)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.119  85.624 170.539 226.315 371.950 626.047

# manhattan distance matrix
dist_man <- dist(Boston, method= "manhattan")

# look at the summary of the distances
summary(dist_man)

##     Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
##    2.016  149.145  279.505  342.899  509.707 1198.265

set.seed(123)

# determine the number of clusters
k_max <- 10

# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Boston, k)$tot.withinss})

# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')

# k-means clustering
km <- kmeans(Boston, centers = 4)

# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)

Optimal number of clusters is when the total WCSS drops radically so according to the plot about 2 clusters.

set.seed(123)

# k-means clustering
km <- kmeans(Boston, centers = 2)

# plot the Boston dataset with clusters
pairs(Boston, col = km$cluster)

# plot the Boston dataset with clusters
pairs(Boston[1:7], col = km$cluster)

# plot the Boston dataset with clusters
pairs(Boston[8:14], col = km$cluster)

Bonus: Perform k-means on the original Boston data with some reasonable number of clusters (> 2). Remember to standardize the dataset. Then perform LDA using the clusters as target classes. Include all the variables in the Boston data in the LDA model. Visualize the results with a biplot (include arrows representing the relationships of the original variables to the LDA solution). Interpret the results. Which variables are the most influential linear separators for the clusters? (0-2 points to compensate any loss of points from the above exercises)

# load the data
data(Boston)

# center and standardize variables
boston_scaled_2 <- scale(Boston) 

# change the object to data frame
boston_scaled_2 <- as.data.frame(boston_scaled_2)

# k-means clustering
km <- kmeans(boston_scaled_2, centers = 4)

boston_scaled_2 <- boston_scaled_2 %>% 
mutate(clusters = c(km$cluster))

# linear discriminant analysis
lda.km <- lda(clusters ~ ., data = boston_scaled_2)

# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

# target classes as numeric

classes <- as.numeric(boston_scaled_2$clusters)

# plot the lda results
plot(lda.km, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.km, myscale = 1)

Super-Bonus: Run the code below for the (scaled) train data that you used to fit the LDA. The code creates a matrix product, which is a projection of the data points.

model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)

## [1] 404  13

dim(lda.fit$scaling)

## [1] 13  3

# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)

Next, install and access the plotly package. Create a 3D plot (cool!) of the columns of the matrix product using the code below.

#install.packages("plotly")
library(plotly)

## 
## Attaching package: 'plotly'

## The following object is masked from 'package:ggplot2':
## 
##     last_plot

## The following object is masked from 'package:MASS':
## 
##     select

## The following object is masked from 'package:stats':
## 
##     filter

## The following object is masked from 'package:graphics':
## 
##     layout

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')

Adjust the code: add argument color as a argument in the plot_ly() function. Set the color to be the crime classes of the train set. Draw another 3D plot where the color is defined by the clusters of the k-means. How do the plots differ? Are there any similarities? (0-3 points to compensate any loss of points from the above exercises)

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)

I did not know how to draw the plot with the color is defined by the clusters of the k-means.